Embedding of $S_3$ in $S_6$

Question

Prove or disprove: There exists an embedding of $S_3$ in $S_6$ as transitive subgroup.

Half proof(attempt): Let $\phi : S_3 \longrightarrow S_6$ that sends cycle $\tau$ in $S_3$ to $S_6$ , For example $\phi (1,2,3)= (1,2,3)$ (but in S_6) , It is a homomorphism. Now, define $X=\{(x_1,...,x_6)|x_4=4,x_5=5,x_6 =6,x_1,x_2,x_3\in \{1,...,3\}\}$ and the action of $Im(\phi)$ $\forall g\in Im(\phi)$ : $$ \sigma.x= (\sigma(x_1),\sigma(x_2), \sigma(x_3), x_4,x_5,x_6) $$

Let $x,y\in X$, Here I'm struggling to show that it is transitive.

Answer 1

$S_3$ acts on itself transitively by multiplication.